RD Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Solutions
Question - 21 : - Solve graphically each of the following systems of linear equations. Also And the coordinates of the points where the lines
meet axis of y.
(i) 2x – 5y + 4 = 0
2x + y – 8 = 0 (C.B.S.E. 2005)
(ii) 3x + 2y = 12
5x – 2y = 4 (C.B.S.E. 2000C)
(iii) 2x + y – 11 = 0
x – y – 1=0 (C.B.S.E. 2000C)
(iv) x + 2y – 7 = 0
2x – y – 4 = 0 (C.B.S.E. 2000C)
(v) 3x + y – 5 = 0
2x – y – 5 = 0 (C.B.S.E. 2002C)
(vi) 2x – y – 5 = 0
x – y – 3 = 0 (C.B.S.E. 2002C)
Answer - 21 : -
(i) 2x – 5y + 4 = 0, 2x – 5y + 4 = 0
2x – 5y + 4 = 0 ⇒ 2x = 5y – 4
⇒ x =
Substituting some different values of y, we get their corresponding values of x as shown here
Now plot the points on the graph and join them
2x + y – 8 = 0 => 2x = 8 – y
x =
Substituting some different values of y, we get their corresponding values of x as shown below
Now join these points and join them
We see that the lines intersect each other at (3, 2)
x = 3, y = 2
These line intersect y-axis at(0, 45) and (0, 8) respectively.
(ii) 3x + 2y = 12, 5x – 2y = 4
3x + 2y = 12
=> 3x = 12 – 2y
x =
Substituting some different values of y, we get their corresponding values of x as shown below
Now plot the points and join them Similarly in 5x – 2y = 4
=> 5x = 4 + 2y
x =
Now join these points and join them
We sec that these lines intersect each other at (2, 3)
x = 2, y= 3
There lines intersect y-axis at (0, 6) and (0, 2) respectively.
(iii) 2x + y – 11 = 0, x – y – 1 = 0
2x + y – 11 = 0 => y = 11 – 2x
Substituting some different values of x, we get their corresponding values of y as shown below:
Now plot the points and join them Similarly in x – y – 1= 0 => x = y + 1
Now plot the points and join them We see that these two lines intersect each othetr at (4, 3) and intersect y-axis at (0, 11) and (0,-1)
(iv) x + 2y – 7 = 0, 2x – y – 4 = 0
x + 2y – 7 = 0
x = 7 – 2y
Substituting some different value of y, we get their corresponding values of x as shown below
Now plot these points and join then Similarly in
2x – y – 4 = 0
y = 2x – 4
Now plot these points and join them We see that these two lines intersect each other at (3, 2)
and these lines intersect y-axis at (0, 7/2) and (0, -4)
(v) 3x + y – 5 = 0, 2x – y – 5 = 0
3x + y – 5= 0
y = 5 – 3x
Substituting some different values of x, we get corresponding values of y as shown below
Now plot these points and join them Similarly in 2x – y – 5 = 6 => y = 2x – 5
Now plot these points and join them We see that these two lines intersect each other at (2, -1)
x = 2, y = 1
and these Lines intersect y-axis at (0, 5) and (0, -5) respectively.
(vi) 2x – y – 5 = 0, x – y – 3 = 0
2x – y – 5 = 0
y = 2x – 5
Substituting some different values of x, we get their corresponding values of y as shown below
Plot the poinfs and join them Similarly in-the equation x – y – 3 = 0 => x =y + 3
Plot these points on the graph and join them we see that these two lines intersect each other at (2, -1)
x = 2, y = 1
and these lines intersect y-axis at (0, -5) and (0, -3)
Solve the following system of linear equations graphically and shade the region between the two lines and x-axis
(i) 2x + 3y = 12, x – y = 1 (C.B.S.E. 2001)
(ii) 3x + 2y – 4 = 0, 2x – 3y – 7 = 0 (C.B.S.E. 2006C)
(iii) 3x + 2y – 11 = 0, 2x – 3y + 10 = 0 (C.B.S.E. 2006C)
Answer - 22 : -
(i) 2x + 3y = 12, x – y = 1
2x + 3y = 12 => 2x = 12 – 3y
=> x =
Substituting some different values of y, we get corresponding values of x as shown below
Now plot the points on the graph and join them. Similarly in the equation
x – y = 1 => x = 1 + y
Now plot the points on the graph and join them
We see the two lines intersect each other at (3, 2) and intersect also x-axis at (6, 0) and 0,0)
The required region has been shaded.
(ii) 3x + 2y – 4 = 0, 2x – 3y – 7 = 0
3x + 2y – 4 = 0
=> 3x = 4 – 2y
x =
Substituting some different values of y, we get corresponding values of x as shown below
Now plot the points on the graph and join them. Similarly in the equation
2x – 3y – 7 = 0
=> 2x = 3y + 7
=> x =
Plot these points and join them
The required region surrounded by these two lines and x-axis has been shaded as shown.
(iii) 3x + 2y – 11 = 0, 2x – 3y + 10 = 0
3x + 2y – 11
=> 3x = 11 – 2y
=> x =
Substituting some different value of y, we get corresponding values of x as shown below
Now plot the points and join them. Similarly in the equation
2x – 3y + 10 = 0
2x = 3y – 10
x =
Now plot the points and join them
The required region surrounded by these two lines and Y-axis has been shaded as shown.
Draw the graphs of the following equations on the same graph paper :
2x + 3y =12
x – y = 1
Find the co-ordinates of the vertices of the triangle formed by the two straight lines and the y-axis. (C.B.S.E. 2001)
Answer - 23 : -
2x + 3y = 12
⇒ 2x = 12 – 3y
x =
Substituting some different values of y, we get corresponding values of x as shown below
Now plot the points on the graph and join them. Similarly in the equation
x – y = 1 => x = y + 1
Now plot the points on the graph and join them
The required region surrounded by these two lines and y-axis has been shaded as shown
Question - 24 : - Draw the graphs of x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and x-axis and shade the triangular area. Calculate the area bounded by these lines and x-axis. (C.B.S.E. 2002)
Answer - 24 : -
x – y + 1 =0, 3x + 2y-12 = 0
x – y + 1 = 0
x = y – 1
Substituting some different values of y, we get so their corresponding values of x as shown below :
Now plot the points and join them Similarly, in the equation
3x + 2y – 12 = 0 => 3x = 12 – 2y
x =
Plot the points on the graph and join them. These two lines intersect each other at (2, 3) and x-axis at (-1, 0) and (4, 0)
Area of the triangle ABC =
x Base x AltitudeSolve graphically the system of linear equations :
4x – 3y + 4 = 0
4x + 3y – 20 = 0
Find the area bounded by these lines and x-axis. (C.B.S.E. 2002)
Answer - 25 : -
4x – 3y + 4 = 0
=> 4x = 3y – 4
=> x =
Substituting some different values of y, we get their corresponding values of x as shown below:
Question - 26 : - Solve the following system of linear equations graphically 3x + y – 11 = 0, x – y – 1 = 0. Shade the region bounded by these lines and y-axis. Also find the area of the region bounded by the these lines and y-axis. (C.B.S.E. 2002C)
Answer - 26 : -
3x + y – 11=0
y = 11 – 3x
Substituting some different values of x, we get their corresponding values of y as shown below :
Now plot these points on the graph and join them Similarly in equation
x – y – 1 = 0
=> x = y + 1
Now plot these points on the graph and join them. We see that these two lines intersect each other at the point (3,2)
x = 3, y = 2
Now shade the region enclosed by these two lines and y-axis
Area of shaded ∆ABC
=
x AC x BD=
x 12 x 3 = 18 sq. units
Question - 27 : - Solve graphically each of the following systems of linear equations. Also find the co-ordinates of the points where the lines meet the axis of x in each system.
(i) 2x + y = 6
x – 2y = -2 (C.B.S.E. 1998)
(ii) 2x – y = 2
4x – y = 8 (C.B.S.E. 1998)
(iii) x + 2y = 5
2x – 3y = -4 (C.B.S.E. 2005)
(iv) 2x + 3y = 8
x – 2y = -3 (C.B.S.E. 2005)
Answer - 27 : -
(i) 2x + y = 6, x – 2y = -2
2x + y = 6
y = 6 – 2x
Substituting some different values of x, we get their corresponding values of y as shown below
Now plot the points and join them Similarly in the equation
x – 2y = -2
=> x = 2y – 2
Now plot the points and join them We see that these two lines intersect each other at (2, 2)
x = 2, y = 2
Here two lines also meet x-axis at (3, 0) and (-2, 0) respectively as shown in the figure.
(ii) 2x – y = 2, 4x – y = 8
2x – y = 2
=> y = 2x – 2
Substituting some different values of x, we get corresponding values of y as shown below:
Now plot the points on the graph and join them Similarly in equation
4x – y = 8
=> y = 4x – 8
Now plot these points and join them We see that these two lines intersect each other at (3, 4)
x = 3, y = 4
These two lines also meet x-axis at (1, 0) and (2,0) respectively as shown in the figure
(iii) x + 2y = 5, 2x – 3y = -4
x + 2y = 5
=> x = 5 – 2y
Substituting some different values of y, we get their corresponding values of x as shown below
Now plot the points on the graph and join them
Similarly in the equation
2x – 3y = 4
=> 2x = 3y – 4
x =
Plot these points and join them
We see that these two lines intersect each other at (1, 2)
x = 1, y = 2
and these two lines meet x-axis at (5, 0) and (-2, 0) respectively as shown in the figure
(iv) 2x + 3y = 8, x – 2y = -3
2x + 3y = 8
=> 2x = 8 – 3y
x =
Substituting some different values of y, we get their corresponding values of x as shown below:
Plot these points on the graph and join them Similarly in equation
x – 2y = -3
x = 2y – 3
Now plot these points and join them We see that these two lines intersect each other at (1, 2)
x = 1, y = 2
and also these lines meet x-axis at (4, 0) and (-3, 0) respectively as shown in the figure
Draw the graphs of the following equations 2x – 3y + 6 = 0
2x + 3y – 18 = 0
y – 2 = 0
Find the vertices of the triangle so obtained. Also, find the area of the triangle.
Answer - 28 : -
2x – 3y + 6 = 0
2x = 3y – 6
x =
Substituting some different values of y, we get their corresponding values of x as shown below:
Now plot these points on the graph and join them
Similarly in the equation 2x + 3y -18 = 0
=> 2x = 18 – 3y
x =
and in equation y – 2 = 0
y = 2
Which is parallel to x-axis on its positive side Now plot the points and join them We see that these lines intersect each other at (3, 4), (6, 2) and (0, 2)
Area of the triangle ABC, so formed
=
x base x altitude=
x BC x AD=
x 6 x 2= 6 sq. units
Question - 29 : - Solve the following system of equations graphically:
2x – 3y + 6 = 0
2x + 3y – 18 = 0
Also find the area of the region bounded by these two lines and y-axis.
Answer - 29 : -
2x – 3y + 6 = 0
2x = 3y – 6
x =
Substituting some different values of y, we get their corresponding values of x as shown below:
Plot these points on the graph and join them Similarly in the equation
2x + 3y – 18 = 0
=> 2x = 18 – 3y
x =
Plot these points on the graph and join them. We see that these two lines intersect each other at (3, 4)
x = 3, y = 4
These lines formed a triangle ABC with the y-axis
Area of ∆ABC =
x base x altitude=
x BC x AD=
x 4 x 3 = 6 Sq. units
Question - 30 : - Solve the following system of linear equations graphically :
4x – 5y – 20 = 0
3x + 5y – 15 = 0
Determine the vertices of the triangle formed by the lines representing the above equation and the y-axis. (C.B.S.E. 2004)
Answer - 30 : -
4x – 5y – 20 = 0
=> 4x = 5y + 20
x =
Substituting some different values of y, we get their corresponding values of x as shown below
Plot these points on the graph and join them Similarly in the equation
3x + 5y – 15 = 0
=> 3x = 15 – 5y
x =
Now plot these points and join them We see that these two lines intersect each other at (5, 0)
x = 5, y = 0
These two lines form a ∆ABC with y-axis whose vertices are A (5, 0), B (0, 3), C (0, -4)